How many moles of Fe3O4 can be produced when 6 mol Fe react with 6 mol O2?

a. how many moles of Fe3O4 can be produced when 6 mol Fe react with 6 mol O2?





the answer will be in moles.





b. how many moles of each reactant are left over?


--- I know that Fe disappears, but I'm not sure how to find how many moles of O2 remain.





I would really appreciate it if you would explain how you arrived at the answer. I'm really trying to understand this, but apparently I'm chemistry illiterate.

How many moles of Fe3O4 can be produced when 6 mol Fe react with 6 mol O2?
No one is chemistry illiterate and the fact that you want to understand how this works proves that, in addition, you also correctly state that the Fe disappears; so there you have it.


Now, to your questions:





a. You are asked to figure out how many moles of Fe3O4 is produced when iron and oxygen react. The first thing to do is set up an equation





Fe + O2 -%26gt; Fe3O4





Next, balance the equation





3Fe + 2O2 -%26gt;Fe3O4





Then, check your equation and make sure its balanced. On the left side you have 3 iron atoms (Fe) and 2 oxygen molecules (O2) or 4 oxygen atoms. On the right side you have 3 iron atoms and 4 oxygens atoms. So they balance.





So, the equation is correct (in fact, the reaction is the oxidation of iron to form rust).





Ok, we now know that we need 3 moles of Fe to react with 2 moles of oxygen to form 1 mole of Fe3O4:





3Fe + 2O2 -%26gt;Fe3O4





Now you're given 6 moles of reactants (the iron and oxygen respectively). You need to find out how much of each you actually use in the reaction. This is known as the limiting reagent problem.





So, from the balanced equation above, you try to figure out how much of the product (Fe3O4) your 6 moles of Fe will yield:


6 moles of Fe x (1 mol of Fe3O4 / 3 moles of Fe) = 2 moles of Fe3O4.





Similarly, you try to figure out how much Fe3O4 your 6 moles of O2 will yield. So, 6 moles of O2 x (1 mol Fe3O4 / 2 moles of O2) = 3 moles of Fe3O4.





So you're limiting reagent (the reactant that runs out first) will be the Fe because you can only get 2 moles of your product (Fe3O4) from your original 6 moles of Fe. This answers the (a) part of your question: 2 moles of Fe3O4 can be generated from the use of 6 moles each of Fe and O2 (and, as you stated in your question, the Fe gets used up).





Now, you know that if Fe was not limiting you, you could make 3 moles of Fe3O4 because you have enough oxygen to do so. Oxygen then is the reactant left over; however,since you ARE limited you have 1 mole of Fe3O4 you can't make.





Now, to find out how many moles of oxygen aren't used up you have to work the problem backwards.





1 mole of Fe3O4 could have been made due to excess oxygen but was not because of your limiting amount of Fe:





Using the original equation:


1 mole of Fe3O4 x (2 moles of O2 / 1 mole of Fe3O4) = 2 moles of O2. And 2 moles of O2 is what you have remaining.





Hope that helps, ask if you have any more questions or if I need to make things more clear.





Best Wishes.